如何在 R 中制作意大利面图? [英] How to make a spaghetti plot in R?
问题描述
我有以下几点:头(数据框):
I have the following: heads(dataframe):
ID Result Days
1 70 0
1 80 23
2 90 15
2 89 30
2 99 40
3 23 24
等等...我想要做的是:用上面的数据创建一个意大利面图.我用的是这个:
ect... what I am trying to do is: Create a spaghetti plot with the above datast. What I use is this:
interaction.plot(dataframe$Days,dataframe$ID,dataframe$Result,xlab="Time",ylab="Results",legend=F)
但没有一条患者线是连续的即使他们应该排很长的队.
interaction.plot(dataframe$Days,dataframe$ID,dataframe$Result,xlab="Time",ylab="Results",legend=F)
but none of the patient lines are continuous even when they were supposed to be a long line.
我也想将上面的数据帧转换成这样的:ID 结果天数
Also I want to convert the above dataframe to something like this: ID Result Days
1 70 0
1 80 23
2 90 0
2 89 15
2 99 25
3 23 0
ect...(我试图取每个 id 的第一个(或最小值),并从零开始计算它们的日期).同样在意大利面图中,我希望所有患者在满足条件时具有相同的颜色,如果不满足条件则使用另一种颜色.
ect... ( I am trying to take the first (or minimum) of each id and have their dating starting from zero and up). Also in the spaghetti plot i want all patients to have the same color IF a condition in met, and another color if the condition is not met.
感谢您的时间和耐心.
推荐答案
这个怎么样,用ggplot2和data.table
How about this, using ggplot2 and data.table
# libs
library(ggplot2)
library(data.table)
# your data
df <- data.table(ID=c(1,1,2,2,2,3),
Result=c(70,80,90,89,99,23),
Days=c(0,23,15,30,40,24))
# adjust each ID to start at day 0, sort
df <- merge(df, df[, list(min_day=min(Days)), by=ID], by='ID')
df[, adj_day:=Days-min_day]
df <- df[order(ID, Days)]
# plot
ggplot(df, aes(x=adj_day, y=Result, color=factor(ID))) +
geom_line() + geom_point() +
theme_bw()
更新后的 data.frame 的内容(实际上是一个 data.table):
Contents of updated data.frame (actually a data.table):
ID Result Days min_day adj_day
1 70 0 0 0
1 80 23 0 23
2 90 15 15 0
2 89 30 15 15
2 99 40 15 25
3 23 24 24 0
您可以使用 scale_color_manual()
You can handle the color coding easily using scale_color_manual()
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